Let $\ell$ be the line perpendicular to $AP$ at $P$. Let $\ell \cap AO=Q$ and we get that $Q$ is the midpoint of $DE$. Easy angle chasing gives us $DB\perp EC$, let their intersection be $R$. Suppose the tangent line at $A$ of $(ABC)$ intersects $BC$ at $K$. Note that $\angle{OAK}=90^{\circ}$. Easy to see that $KA=KT$ and so $\angle{KMA}=90^{\circ}$. Then, we'll proof that $K,D,E$ collinear. This is true by considering radical axises of circles $(APQ),(ABC),$ and $(PBC)$, we get Radical axis of $(APQ)$ and $(ABC)$ is the tangent line at $A$ of $(ABC)$ (by angle chasing.) Radical axis of $(ABC)$ and $(PBC)$ is $BC$. Radical axis of $(APQ)$ and $(PBC)$ is $PQ=\ell$ if one can show that $Q$ lie on $(BPC)$. For the last line, we have to show that $Q$ lie on $(BPC)$, this is true if one can prove the following lemma: Lemma wrote: Given a right triangle $ERD$ with $\angle{ERD}=90^{\circ}$. Let $K$ be the midpoint of $ED$ and $P$ be an arbitrary point on the side $DE$. Draw perpendicular bisector of $DP$ and $EP$ intersects the sides $DR$ and $ER$ at $B$ and $C$, respectively. Then $P,Q,C,B$ concyclic. Which I'm too lazy to find synthetic proof. Now, from $K,D,E$ collinear, we get that $\angle{KPA}=90^{\circ}$ and then $K,M,P,A$ concyclic. This finish the problem by angle chasing that $\angle{OAP}=\angle{PKA}=\angle{PMA}$.

Let $K = DE\cap BC$ and $N$ is the midpoint of $DE$. The key step is to prove that $B, C, P, N$ are concyclic. To do that, let $X, Y$ be the projections from $B, C$ to $DE$ respectively, which are midpoints of $PD, PE$ respectively. Since $\angle BPC = 90^{\circ}$, triangles $PBX$ and $CPY$ are similar. Moreover, simple side-chasing gives $XN=YP$ thus $$\frac{XN}{XB} = \frac{PY}{BX} = \frac{PX}{CY} = \frac{NY}{YC}$$or triangles $NBX$ and $CNY$ are similar, which implies $\angle BNC=90^{\circ}$ as claimed. Since $K$ is radical center of $\odot(APN), \odot(BCPN), \odot(ABC)$, it lies on radical axis of $\odot(APN), \odot(ABC)$. However, these circles are obviously tangent so $AK$ is tangent to $\odot(ABC)$. Finally, recall that $KA=KT$ so $\angle AMK = 90^{\circ} = \angle APK$. Hence $\odot(AMP)$ is circle with diameter $AK$, which is obviously orthogonal to $\odot(ABC)$, implying the conclusion.

CSEMO 2018 Grade 10 Q3 wrote: Let $O$ be the circumcenter of acute $\triangle ABC$($AB<AC$), the angle bisector of $\angle BAC$ meets $BC$ at $T$ and $M$ is the midpoint of $AT$. Point $P$ lies inside $\triangle ABC$ such that $PB\perp PC$. $D,E$ distinct from $P$ lies on the perpendicular to $AP$ through $P$ such that $BD=BP, CE=CP$. If $AO$ bisects segment $DE$, prove that $AO$ is tangent to the circumcircle of $\triangle AMP$. Great problem. Let $F$ be the midpoint of $DE$, $I$ be the intersection of $BD$ and $CE$, and $J$ be the intersection of $BC$ and $DE$. Claim 01. $BPFC$ is cyclic. Proof. Notice that $DB \perp CE$. Since $F$ is the midpoint of $DE$, we have \[ \measuredangle FIB \equiv \measuredangle FID = \measuredangle IDF \equiv \measuredangle BDP = \measuredangle DPB = \measuredangle FPB \]Therefore $BPFI$ is cyclic. Similarly, we get $BFCI$ being cyclic. Since $\measuredangle BIC = \measuredangle BPC = 90^{\circ}$ as well, we have $BPCI$ is cyclic. Therefore, we conclude that $BPFC$ is cyclic. Claim 02. $AJ$ tangent to $(ABC)$. Proof. By radical axis theorem on $(ABC), (APF), (BPFC)$. We have $PF \cap BC = J$ lie on the radical axis of both $(ABC)$ and $(APF)$. However, $(APF)$ has diameter $AF \equiv AO$, and therefore these circles are tangent to each other at $A$. Hence, $AJ$ is the radical axis of these circle, proving that $AJ$ is tangent to $(ABC)$. Claim 03. $APMJ$ is cyclic. Proof. Since $JA$ is tangent to circle and $AT$ is the angle bisector of $\angle BAC$, we have $JA = JT$. Since $M$ is the midpoint of $AT$, we have $JM \perp AT$. This forces \[ \measuredangle JMA = 90^{\circ} = \measuredangle JPA \]proving $APMJ$ is cyclic. Claim 04. $AO$ is tangent to $(AMP)$. Proof. Notice that $AJ$ is tangent to $(ABC)$, and therefore $AJ \perp AO$. Since $(APMJ)$ has diameter $AJ$ and $(ABC)$ has radius $AO$. We conclude that $(APMJ)$ and $(ABC)$ are orthogonal, forcing $AO$ is tangent to $(AMP)$.